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<p><dfn class="terminology">Solution</dfn> The corresponding homogeneous equation is</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq3_35.html ./knowl/eq3_35.html ./knowl/eq3_36.html">
\begin{equation}
x^2 y^{\prime \prime}-2 y=0.\tag{3.7.14}
\end{equation}
</div>
<p class="continuation">Observation: coefficients are <span class="process-math">\(x^2, 2\text{,}\)</span> and we may seek a solution of the form</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq3_35.html ./knowl/eq3_35.html ./knowl/eq3_36.html">
\begin{equation*}
y=A x^2+B x+C.
\end{equation*}
</div>
<p class="continuation">Taking the solution into (<a href="" class="xref" data-knowl="./knowl/eq3_35.html" title="Equation 3.7.14">(3.7.14)</a>),</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq3_35.html ./knowl/eq3_35.html ./knowl/eq3_36.html">
\begin{equation*}
x^2 2A-2 A x^2-2 B x-2 C=0 \to B=C=0.
\end{equation*}
</div>
<p class="continuation">Thus <span class="process-math">\(y_1=A x^2\)</span> is a solution and we may set <span class="process-math">\(A=1\text{.}\)</span> To find another solution, we use the method of reduction of order and let</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq3_35.html ./knowl/eq3_35.html ./knowl/eq3_36.html">
\begin{equation*}
y_2=v(x) y_1=v(x) x^2.
\end{equation*}
</div>
<p class="continuation">Substituting it into (<a href="" class="xref" data-knowl="./knowl/eq3_35.html" title="Equation 3.7.14">(3.7.14)</a>),</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq3_35.html ./knowl/eq3_35.html ./knowl/eq3_36.html">
\begin{equation}
x^2 (v^{\prime \prime} x^2+4 x v^{\prime}+2 v)-2 v x^2=0.\tag{3.7.15}
\end{equation}
</div>
<p class="continuation">Setting <span class="process-math">\(u=v^{\prime}\text{,}\)</span> (<a href="" class="xref" data-knowl="./knowl/eq3_36.html" title="Equation 3.7.15">(3.7.15)</a>) becomes</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq3_35.html ./knowl/eq3_35.html ./knowl/eq3_36.html">
\begin{equation*}
\begin{aligned}
&amp;x^4 \frac{\textrm{d} u}{\textrm{d} x}+ 4 x^3 u=0 \to \frac{\textrm{d} u}{u}=-\frac{4}{x} \textrm{d} x\\
&amp;\ln |u|=-4 \ln |x|+k_1 \to u=\frac{k_1}{x^4}=\frac{-3}{x^4},
\end{aligned}
\end{equation*}
</div>
<p class="continuation">where the arbitrary constant <span class="process-math">\(k_1\)</span> is taken to be <span class="process-math">\(-3\text{.}\)</span> Then</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq3_35.html ./knowl/eq3_35.html ./knowl/eq3_36.html">
\begin{equation*}
v^{\prime}=u=\frac{-3}{x^4} \to v=\frac{1}{x^3}+k_2=\frac{1}{x^3},
\end{equation*}
</div>
<p class="continuation">where the arbitrary constant <span class="process-math">\(k_2\)</span> is set to be <span class="process-math">\(0\text{.}\)</span> Thus</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq3_35.html ./knowl/eq3_35.html ./knowl/eq3_36.html">
\begin{equation*}
y_2=v(x) y_1=\frac{1}{x}.
\end{equation*}
</div>
<p class="continuation">The general solution for the homogeneous equation is</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq3_35.html ./knowl/eq3_35.html ./knowl/eq3_36.html">
\begin{equation*}
y=C_1 x^2+C_2 \frac{1}{x}.
\end{equation*}
</div>
<span class="incontext"><a href="sec3_7.html#p-143" class="internal">in-context</a></span>
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